Question

Hydrocarbon mixtures are used as fuels. (a) How many grams of \(\mathrm{CO}_{2}(g)\) are produced by the combustion of \(200 . \mathrm{g}\) of a mixture that is \(25.0 \% \mathrm{CH}_{4}\) and \(75.0 \% \mathrm{C}_{3} \mathrm{H}_{8}\) by mass? (b) A \(252-\mathrm{g}\) gaseous mixture of \(\mathrm{CH}_{4}\) and \(\mathrm{C}_{3} \mathrm{H}_{8}\) burns in excess \(\mathrm{O}_{2},\) and \(748 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) gas is collected. What is the mass \(\%\) of \(\mathrm{CH}_{4}\) in the mixture?

Short answer

(a) 586.12 g \(\text{CO}_{2}\), (b) 19% \(\text{CH}_{4}\).

Step-by-step solution

Calculate the mass of each component in the mixture (Part a)

First, find the mass of each hydrocarbon in the 200 g mixture. Since the mixture is 25.0% \(\text{CH}_{4}\) and 75.0% \(\text{C}_{3}\text{H}_{8}\): - Mass of \(\text{CH}_{4}\) = 200 g \times 0.25 = 50 \text{ g} - Mass of \(\text{C}_{3}\text{H}_{8}\) = 200 g \times 0.75 = 150 \text{ g}

Write balanced combustion reactions

Next, write the balanced chemical equations for the combustion of \(\text{CH}_{4}\) and \(\text{C}_{3}\text{H}_{8}\): - \(\text{CH}_{4} + 2 \text{O}_{2} \rightarrow \text{CO}_{2} + 2 \text{H}_{2}\text{O}\) - \(\text{C}_{3}\text{H}_{8} + 5 \text{O}_{2} \rightarrow 3 \text{CO}_{2} + 4 \text{H}_{2}\text{O}\)

Calculate moles of each reactant

Calculate the number of moles for each hydrocarbon: - Moles of \(\text{CH}_{4}\): \(\frac{50 \text{ g}}{16.04 \text{ g/mol}} = 3.12 \text{ mol}\) - Moles of \(\text{C}_{3}\text{H}_{8}\): \(\frac{150 \text{ g}}{44.10 \text{ g/mol}} = 3.40 \text{ mol}\)

Calculate moles of \(\text{CO}_{2}\) produced

Find the amount of \(\text{CO}_{2}\) produced in each reaction: - From \(\text{CH}_{4}\): \(\text{CO}_{2}\): \(\text{3.12 mol} \times 1 = 3.12 \text{ mol}\) - From \(\text{C}_{3}\text{H}_{8}\): \(\text{CO}_{2}\): \(\text{3.4 mol} \times 3 = 10.2 \text{ mol}\)

Convert moles of \(\text{CO}_{2}\) to grams

Calculate the total mass of \(\text{CO}_{2}\) produced by adding the contributions from both hydrocarbons and using the molar mass of \(\text{CO}_{2}\) (44.01 g/mol): - Total moles of \(\text{CO}_{2}\) = 3.12 mol + 10.2 mol = 13.32 mol - Mass of \(\text{CO}_{2}\): \(\text{13.32 mol} \times 44.01 \text{ g/mol} = 586.12 \text{ g}/2. Convert to whole number: \approx{586.12}\text{g}\)

Write the balanced combustion reactions (Part b)

Refer to the balanced equations previously used: \(\text{CH}_{4} + 2 \text{O}_{2} \rightarrow \text{CO}_{2} + 2 \text{H}_{2}\text{O}\) \(\text{C}_{3}\text{H}_{8} + 5 \text{O}_{2} \rightarrow 3 \text{CO}_{2} + 4 \text{H}_{2}\text{O}\)

Set up systems of equations

Let x and y be the masses of \(\text{CH}_{4}\) and \(\text{C}_{3}\text{H}_{8}\) in the mixture: \[ x + y = 252 \text{ g} \] \[ \frac{x}{16.04} + 3\frac{y}{44.10} = \frac{748 \text{ g}}{44.01} = 17 \text{ mol} \]

Solve the system of equations

Solve for x and y: Using substitution or elimination, substitute values and find: - x = 48 g - y = 204 g Finally, calculate the mass %: Mass % of \(\text{CH}_{4}\): \[ \frac{48}{252} \times 100 = 19\text{%} \]

Key concepts

chemical reactions

Chemical reactions are processes where substances (reactants) interact to form new substances (products). In our exercise, we are looking at the combustion of hydrocarbons, where methane \((\text{CH}_{4})\) and propane \((\text{C}_{3}\text{H}_{8})\) combust (react with oxygen) to produce carbon dioxide \((\text{CO}_{2})\) and water \((\text{H}_{2}\text{O})\). This type of reaction is called a combustion reaction. Here's why:

• *Combustion reactions* involve a fuel (hydrocarbon) reacting with oxygen.
• These reactions release energy, typically in the form of heat and light.
• The reactants get transformed into different products, often involving carbon dioxide and water.

Understanding the principles of chemical reactions helps us predict the products and quantities formed from given starting materials.

stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It is based on the law of conservation of mass and takes into account the relative amounts (ratios) of substances involved:

• Calculate the amount of products formed from given reactants.
• Use balanced chemical equations to find these ratios.
• Determine the limiting reagent, which is the reactant that gets completely used up first, limiting the amount of product formed.

For example, in the combustion of methane \((\text{CH}_{4})\), we used the equation \(\text{CH}_4 + 2 \text{O}_2 \rightarrow \text{CO}_2 + 2 \text{H}_2\text{O}\). Using stoichiometry, we found out how many moles of \(\text{CO}_{2}\) are produced from a known amount of \(\text{CH}_{4}\).

With stoichiometry, we also determined the mass percent of a component in a mixture by understanding the relationship between the mass and the moles of each substance.

molar mass

The molar mass is the mass of one mole of a substance, measured in grams per mole (g/mol). It is crucial for converting between grams of a substance and the number of moles. To find molar mass, add up the atomic masses of all atoms in the molecule:

• *Methane* \((\text{CH}_{4})\): 1 C (12.01 g/mol) + 4 H (1.01 g/mol x 4) = 16.04 g/mol
• *Propane* \((\text{C}_{3}\text{H}_{8})\): 3 C (12.01 g/mol x 3) + 8 H (1.01 g/mol x 8) = 44.10 g/mol

Knowing the molar mass, we were able to convert the given mass of hydrocarbons into moles. For example:
\(\text{Moles of CH}_{4} = \frac{50 \text{ g}}{16.04 \text{ g/mol}} = 3.12 \text{ mol}\).

The conversion is crucial for using balanced chemical equations in stoichiometry, as stoichiometric coefficients are generally given in moles.

balanced equations

Balanced equations are essential because they ensure that the same number of atoms of each element is present on both sides of the reaction – this reflects the law of conservation of mass. Here’s how to balance equations and why it’s important:

• Identify the reactants and products.
• Adjust coefficients to ensure the number of each type of atom is the same on both sides.
• Check to make sure the equation is balanced by recounting the atoms.

Take the combustion of propane as an example: \(\text{C}_{3}\text{H}_{8} + 5 \text{O}_{2} \rightarrow 3 \text{CO}_{2} + 4 \text{H}_{2}\text{O}\). Here, one molecule of propane reacts with five molecules of oxygen to produce three molecules of carbon dioxide and four molecules of water.

Balancing ensures we can accurately predict the amounts of products formed, which is essential for practical applications like determining how much fuel to use or the environmental impact of emissions.